$g(n) = 4n^{2}-2n+2-5(f(n))$ $h(x) = 2x^{2}-6x-2-4(f(x))$ $f(x) = 6x^{3}+4x^{2}-7x$ $ f(g(0)) = {?} $
First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = 4(0^{2})+(-2)(0)+2-5(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 6(0^{3})+4(0^{2})+(-7)(0)$ $f(0) = 0$ That means $g(0) = 4(0^{2})+(-2)(0)+2+(-5)(0)$ $g(0) = 2$ Now we know that $g(0) = 2$ . Let's solve for $f(g(0))$ , which is $f(2)$ $f(2) = 6(2^{3})+4(2^{2})+(-7)(2)$ $f(2) = 50$